The heart of particle physics
How do we predict?
The major phenomena we observe in particle physics are Decays and Collisions.
Decay 衰變
Everything which is not forbidden is allowed (a principle of English Law)
That which is not explicitly forbidden is guaranteed to occur.
Every particle will decay if it get a chance.
Relativity allows particles to decay by transforming mass into energy.
Heavier particles usually don’t exist in nature.
They decay soon after they are produced.
Decay chain will continue until it can decay no more,
forbidden usually by conservation law (symmetry).
The only five stable particles in nature
Neutron
Proton
Electron
Photon
Neutrino
但中子是會衰變的！
β Decay

n  p  e 
基本粒子是會消失而形成其他粒子的！
在交互作用的交點，粒子是會消失的。
d
W

u

W



π 子的衰變




    e   
Bubble Chamber picture of pions
衰變是一個不確定的機率過程。
我們無法預測單一一顆中子何時衰變，只能預測衰變發生的機率。
單位時間內的衰變機率：衰變率 Γ
如果是處理一大群中子，知道衰變的機率就足夠了：
單一一顆中子的衰變機率即對應一大群中子的衰變分布。
dN
 N
Γ即是一個中子每秒衰變的機率!
dt
N  N 0e
 t
不衰變的機率即不衰變的粒子數
隨時間以指數遞減
β Decay

n  p  e 
衰變的產物可以有連續分布的可能的動量。
物理只能預測衰變為某一個動量組合的機率密度。
衰變率（機率密度）是產物粒子的能量及動量的函數。
    p p , p e , p    E e , E p ,  e 
甚至衰變產物也有一個以上的可能。
不同的可能稱為 Channel
每一個 Channel 就對應一個衰變率。
總衰變率就是所有衰變率的和。
N  N 0e
 t
 N 0e

t

未衰變的粒子數在 t > τ 後就很少了。
τ 稱為Life Time 生命期



  
π± 的衰變是透過弱作用，生命期約 10-8s，
  2
0
π0 的衰變是透過電磁作用，生命期約 10-15s，
若粒子透過強作用衰變，生命期約只有10-23s
這樣的粒子即使產生，都無法在實驗室內看到他的痕跡。
這樣的粒子會以共振曲線的形式出現在其衰變產物的的散射分布上




  p   p


  p 





 p

強作用衰變
散射率對質心能量的分布


 p

這個過程會對以上散射率增加一個共振。
在質心能量等於共振態質量時，會被加強。

1
E  m 
2


2
4
Breit-Wigner Resonance
強衰變的粒子會以共振曲線出現在其衰變產物的的散射分布上

1
E  m 
2


2
4
共振曲線中心即粒子質量。
共振曲線寬度即衰變率。
衰變率 Γ又稱為 Decay Width
Collision
粒子束越強，單位面積粒子數越多，反應發生的次數越多！
Event Rate
dN  L
通量 L Luminosity(亮度)為單位時間通過單位面積的入射粒子數：
∆𝑡 時間內通過面積 A 的粒子數
N    A  v  t
通量 L 為單位時間通過單位面積的粒子數：
L
N
t  A
  v
達到的 Luminosity 是加速器效率的度量
粒子束越強，單位面積粒子數越多，反應發生的次數越多！
Event Rate
定義 dσ
dN  L
d 
dN
L
dσ 是一個面積。稱為散射截面 Scattering Cross Section。
dσ 是一個與粒子束強弱無關的量，只由該反應決定。
dσ 是此反應的內在性質。
dσ 一計算出來，就可以用在所有的實驗。
dσ 之於 dN，就如同比熱之於熱容量。
在古典散射𝑑𝜎真的是截面積
Classical Scattering
Impact parameter b 與散射角度 θ 有一對一對應
散射角在 θ 與 θ+dθ 間的粒子，其 b 必定在對應的 b 與 b+db之間
通過左方此一截面的粒子，將散射進入對應的散射角 𝜃 及𝜃 + 𝑑𝜃 之間
故散射角為 𝜃 的散射粒子數：
dN  L  d 
d   2  b ( )  db  2  b ( ) 
db
d
 d
dN  L  d 
d 
dN
散射截面 Scattering Cross Section
L
d 
dN
L
L
粒子物理的粒子已不再有特定軌跡了。古典的計算已不適用。但舊稱仍沿用。
在此定義下，dσ 依舊是此反應的內在性質。
7x10 12 eV
10 34 cm -2 s -1
2835
10 11
Beam Energy
Luminosity
Bunches/Beam
Protons/Bunch
7.5 m (25 ns)
Bunch Crossing
Proton Collisions
4 10
7
7 TeV Proton Proton
colliding beams
Hz
10 9 Hz
e
Parton Collisions
New Particle Production
(Higgs, SUSY, ....)
10 -5 Hz
p
H
1
µ
-
p
~
q
p
Selection of 1 event in 10,000,000,000,000
q
p
~
q

q
µ-
-
~
g
Z
µ+
e
q
µ+
Z
-
~
 20



~ 0
1
Event recorded with the CMS detector in 2012 at a proton-proton
centre-of-mass energy of 8 TeV. The event shows characteristics
expected from the decay of the SM Higgs boson to a pair of Z bosons,
one of which subsequently decays to a pair of electrons (green lines
and green towers) and the other Z decays to a pair of muons (red lines).
L  v 
N AN B
f
A
dσ 是一個垂直於射向的面積。
在沿射向的羅倫茲變換下是不變的！
Fixed Target實驗的 dσ 與 Colliding Beam實驗一樣。
生成的粒子可以觀察到軌跡
因此生成的粒子可以看成波包。
Δx
粒子的動量波函數有一個分布！
衰變率必須對動量波函數積分：
 
 dp   ( p )  b ( p )
2
Δk
假設我們對粒子位置並未太精密測量
Δp 就不會太大
我們可以以 p0 的衰變率來近似！
   ( p0 )
x  k 
1
2
討論時會近似使粒子都具有一個特定動量，而忽略動量分布！
先將一定會出現的 Factors 從衰變率及截面中提出來。
這一些 Factors 與作用的細節無關，
只和入射粒子及產生粒子的數量與身分有關！
Phase Space Factors
Particle decay: 1( p1 )  2 ( p 2 )  3 ( p 3 )    n ( p n )



衰變後第 i 個粒子的動量在  p i , p i  d p i  之間的衰變率記為 dΓ
py
dpy
dpx
px
衰變率dΓ應與所產生粒子的動量所佔相空間的區間大小成正比
d  M
2
n
  dp jx dp jy dp
jz
 M
2
j2
But this integral is not Lorentz invariant!
n
3 
 d p j
j2
Dirac 𝛿 function
 x  a   0
if x  a

 dx    x  a   1


 dx    x  a   f  x   f  a 

In an integration, enforce the equation that x = a.
 x  a
2


只有在 x = ± a 不為零。   x 2  a 2   c1  x  a   c 2 ( x  a ) 

2
 dx  x  a
2
2
2
1
0

 dx   x  a
2
2

2
  1   2 x  dx  x
0

2
a
2
0

 2 x  dx
   2 x  dx   x
2
a
2

0
c1  x  a   1
2 a  c1  1
c 2  x  a    1
 2a  c2  1
0
0
 2 x  dx

 x  a
2
2

1
2a
  x  a   ( x  a ) 
  f ( x) 

j
  kx  
1
k
 x 
1
f '(x j )
 x  x j  f ( x j )  0
We make it more complicated by allowing an indefinite p0
integration and then fixing it by requiring the on-shell condition:
4

d p
2 
  2  p  m c
4
2
2
2
   p 
0
But in this form, we can be sure it is Lorentz invariant!
We can perform the p0 integration to recover the 3 space form.
 p  m c
2
2
 x  a
2

 p

0 2
2


 p
2
    p 
0 2
1
2a
2

 p
2
m c
2
m c
2
2
  p
0

d p
3
1
2
2
m c
2

  x  a    ( x  a ) 

2
2 3

1
  p 
1
2

p
2

p
2
0
m c
2
2


p
2
2 2 
m c 


d p
3
2 E 2
3 This is Lorentz invariant.
d  M
2
n


d pj
3
我們從衰變率再拉出這個 Factor
使積分是羅倫茲不變
j2
n

j2
d  M
2
n

j2
 2  4  4  p1 
d  M
2
 2
1
n
1

2E j
1

j2

d pj
2
 2 2
2
pj c  mj
3
2 E j 2
3
No matter what, the overall 4momenta are conserved!
p 2  p 3   p n  從衰變率再拉出一個執行動量守恆
的 δ function
4  4  p1 
n
p2  p3   pn  
j2
1
3 
d pj
2 E j 2
3
All the factors are Lorentz Invariant. But is M 2 Lorentz invariant?
d  M
2
 2
4  4  p1 
n
p2  p3   pn  
j2
1
3 
d pj
1
2 E j 2
E1
3
t 
But M 2 is not Lorentz invariant since Γ is not.
For a particle, Γ transforms like 1/t1. t1 transforms like E1.
E 
我們可以從
d 
1
E1
M
2
 2
   p1 
4
4
M2 再提出一個
1/ E1
n
p2  p3   pn  
Now we can be sure M 2 is Lorentz invariant.
It’s called Feynman Amplitude.
The Lorentz Invariance makes M 2 simple.
j2
1

2

v 
 1  2 
c 

mc
1
3 
d pj
2 E j 2
3
2
v
2
c
2
d 
1
E1
M
2
 2
4  4  p1 
n
p2  p3   pn  
j2
1

d pj
3
2 E j 2
3
Dynamic Factors 由交互作用的細節決定。
Kinematic factors只和入射粒子及產生粒子的數量與身分有關！
We separate the kinematics and dynamics in such an elegant way
that the still dynamic part M 2 is Lorentz invariant.
M2
Now we apply this to pion two photon decay.
Choose the rest frame of pion:
d 
1
E1
M
2
 2
  4  p1 
4
p2  p3  
0

p1  0
E1  m 
1
  2
3 
d p2
2 E 2  2
3
1
3 
d p3
2 E 3 2
3
Total Decay Rate:
 dx  f ( x )    x  a  
f (a )
2

3 
d p  p  d p  sin   d   d 
式子與角度無關！
若生成粒子有質量：見課本推導。
注意M 2是沒有因次的。
Two body scattering:
d  M
2
 2
4  4  p1 
1 2  3  n
n
p2  p3   pn  
j3
1

d pj
3
2 E j 2
3
M 2 is Lorentz invariant since dσ almost is.
If we consider only Lorentz transformation along the 1-2
colliding axis, the cross section dσ is invariant!
But we do want to pull out a Lorentz invariant factor that reflects
the inverse luminosity that must appear in cross section:
dN
d 
L
4
 p1  p 2 2   m 1 m 2 2
Lorentz invariant factor that becomes luminosity in rest frame:
In the rest frame of particle 2
4
 E1  m 2 
2
 m1 m 2   4 m 2
2
E2  m2
E m
2
1
2
1

p2  0

 4 m 2 p1  E 2  E 1  v1   2   1 v1
 
1
x

1
v
1  
c
 E
2
d 
M
2
 2
   p1 
4
4
n
p2  p3   pn  
j3
1
4
 p1  p 2 
2
 m1 m 2 
2
1
d 
4
 p1  p 2 2   m 1 m 2 2
1

d pj
3
2 E j 2
3
我們從截面再拉出這個一定要出現
的羅倫茲不變的 Luminosity。
M
2
 2
4  4  p1 
n
p2  p3   pn  
M 2 is Lorentz invariant.
It’s called Feynman Amplitude.
The Lorentz Invariance makes M 2 simple.
j3
1
3 
d pj
2 E j 2
3
Two body scattering in CM
Carry out the p4 integration
Enforce the 3 momentum conservation
散射截面的角度分布
注意M 2是沒有因次的。
Two body scattering in CM
注意M 2是沒有因次的。
如果生成粒子與入射粒子質量相等
d
d

M
2
2
E CM


pi  p f
Feynman Rules
To evaluate the Lorentz Invariant Feynman Amplitude M
Components of Feynman Diagrams
pi
External Lines
qi
Internal Lines
Determined by
Particle Content
(Masses and spins)
Every line contains a 4-momentum.
Vertex
Interactions
Every component corresponds to a specific factor!
pi
External Lines
1 (For spinless particles)
Its momentum is on-shell. 𝑝2 = 𝑚2
qi
Internal Lines
i
q j  m j  i
2
2
propagator
Its momentum is not on-shell.
Vertex
-ig
Coupling Constant

 2  4  4  

i

pi 

Momentum Conservation
Draw all diagrams with the appropriate external lines, matching the
incoming and outgoing particles with momenta fixed by the experiment.
Multiply all the factors!
Integrate over all internal momenta
1
 2  4
4
d qi
Take out an overall momentum conservation.
 2  4  4  p1 
That’s it! The result is -iM.
p2  p3   pn 
It’s so simple.
The Lorentz Invariance is explicit at every step.
A toy ABC model
There are 3 scalar particle with masses mA, mB ,mC
pi
𝑝𝐴2 = 𝑚𝐴2 etc.
1
External Lines
qi
i
Internal Lines
q j  m j  i
2
2
Lines for each kind of particle with appropriate masses.
C
k2
Vertex
A
k1
-ig
k3
 2  4  4  k 1  k 2  k 3 
B
The configuration of the vertex determine the interaction of the model.
Vertices for real Interactions
e, q
e, q
Particle A decay
A BC
 iM   ig   2 
4   4  p1 
p2  p3 
Take out overall momentum conservation.
M  g
The Feynman Amplitude is just a constant!
There are other possible diagrams, in fact infinite number of them.
Each extra vertex carries an extra factor of –ig, which is small.
Consider first the diagrams with the fewest number of vertices.
Feynman Rule is a perturbation theory.
Lifetime of A
A BC
M  g
As an example, consider that a particle is pion and B,C particles are photons.
Assume that the interaction of 𝜋 − 𝛾 − 𝛾 has a coupling constant g.
For  0  2  m B  m C  0

m
m
p  A  
2
2
 
g
2
16   cm 
Scattering
It takes at least two vertices to draw a diagram with appropriate external lines.
The leading order diagram:
It’s similar to electron-electron scattering!
One propagator
and two coupling constants
  ig 2
Carrying out the momentum integration will enforce
momentum conservation at one of the vertex:
dq
4
 2 
4
4


2

 q 
4
p4  p2 
q  p4  p2
Take out overall momentum conservation.
The overall momentum conservation is always there.
 2  4  4  q 
p 4  p 2   2
4  4  p1 
p3  p4  p2 
The remaining momentum conservations can be enforced by
immediately carrying out the integration in the beginning!
dq
p4  p2
4
 2 
4
4


q 
2


4
p4  p2 
The result can be written down right away!
p4  p2
After momentum conservation is enforced, the momentum of internal particle
c is p 4  p 2
2 2
It does not satisfy momentum mass relation of a particle!  p 4  p 2   m c c
2
The internal particle is not a real particle, it’s virtual.
p4  p2
It does not satisfy momentum mass relation of a particle!  p
 p2   mc c
2
4
The internal particle is not a real particle, it’s virtual.
In a sense, the mass relation is only for particles when we “see” them!
Internal lines are by definition “unseen” or “unobserved”.
It’s more like:
p4  p2
It’s more a propagation of fields than particles!
2
2
p4  p2
Field fluctuation propagation can only proceed forward
that is along “time“ from past to future.
This diagram is actually a Fourier Transformation into
momentum space of the spacetime diagram.
The vertices can happen at any spacetime location xμ and the
“location” it happens need to be integrated over. After all
you do not measure where interactions happen and just like
double slit interference you need to sum over all possibilities.
For those amplitude where
time 2 is ahead of time 1,
propagation is from 2 to 1.
For those amplitude where
time 1 is ahead of time 2,
propagation is from 1 to 2.
B
B
B
2
B
1
1
C
C
A
A
p4  p2
2
A
A
is actually the sum of the above two diagrams!
Feynman diagram in momentum space is
much simpler than in space-time!
p4  p2
p3  p 2
This is very similar to electron-electron scattering.
Scattering
To the leading order, there could be more than one diagrams!
In the diagrams, the B lines in some vertices are incoming particle.
The striking lesson: Lines in a vertices can be either
outgoing or incoming, depending on their p0.
p0 for an observed particle is always positive.
So this vertex diagram is actually 8 diagrams put
together! That’s the simplicity of Feynman diagram.
If all the momenta in the diagrams can be determined through
momentum conservation, the diagram has no loop and is hence called
tree diagram!
p4  p2
If there is a loop in the diagram, some internal momentum is not
fixed and has to be integrated over! These are called loop diagram.
將所有的動量守恆事先執行
q
p1  p 3
p1  p 3
p1  p 3  q
若有 loop 就會有某些動量無法確定
無法確定的動量必須積分。

Oops! Loops are infinite!
Renormalization
p4  p2
  
Cut off the
momentum q at Λ

m  m
2
c
2
c , Physical
 m
2
δm中的 ln Λ正好使 Tree 圖與 Loop圖的 ln Λ抵消
  
結果會是有限的！

 ln 
Mass Renormalization
q
p
p
p
p
p
p
p
q p
i
p m
2
i
2
c
p m
2
2
c

i
i
p m
2
p m
2
2
c
2
c







i
1
i
1




 2

2
2
2
2
2


i
p  mc
p  m c  p  m c  i

2 
1  2
2
2
p

m
p

m
c 

c

p m
2
2
c

i
p  mc
2
i
p  m c , Physical
2
2


i

 2
2

p  m c  i


m c  m c , Physical   m
2
2
i
2
i   m
2
2
 ln 
δm中的 ln Λ正好與 Σ 的 ln Λ抵消
2
c 的原始質量是無限大，加上無限大的修正，量到的質量是有限的 m c , Physical
Charge Renormalization
 ln 
g
g  g Physical   g
g Physical
原始耦合常數是無限大，加上無限大的修正，量到的有限的耦合常數 g Physical
All the infinities can be cancelled out by a
finite numbers of parameter renormalization.
Schrodinger Wave Equation
He started with the energy-momentum relation for a particle
Expecting them to act on plane waves
e
 iEt
e
 
ip r
he made the quantum mechanical replacement:
How about a relativistic particle？
e
 ipx
The Quantum mechanical replacement can be made in
a covariant form. Just remember the plane wave can
 
 ipx
be written in a covariant form:  iEt i p  r
e
e
e
As a wave equation, it does not work.
It doesn’t have a conserved probability density.
It has negative energy solutions.
Plane wave solutions for KG Eq.
 ( x)  a  e
p0  a  e
2

2
p0  p
p
 ipx
2
 
0
 ip t  i p  x
 a e
 ipx
2
 ipx
2
 ipx
 p a e
 m a e
m
2
p m c 0
2
2
2
 E
0
Time dependence can be determined.
There are two solutions for each 3 momentum p (one for +E and one for –E )
 ( x) 
 a  e
 ipx

 b e
ipx

p
It has negative energy solutions.
   a  e

p
 
 iEt  i p  x

 b e
 
iEt  i p  x

The proper way to interpret KG equation is it is not a Wavefunction
Equation but actually a Field equation just like Maxwell’s Equations.
Plane wave solutions just corresponds to Plane Waves.
It’s natural for plane waves to contain negative frequency components.
Expansion of the KG Field by plane：
 ( x) 
 a  e
 ipx

 b e
ipx

p
   a  e
 
 iEt  i p  x

p
If Φ is a real function, the coefficients are related:

 
 ( x) 
 a  e

p
 ipx

 a e
ipx


 b e
 
iEt  i p  x

Add a source to the equation:

     m   j( x)
2
We can solve it by Green Function.
   G ( x , x ' )  m  ( x , x ' )    x  x '

2
G is the solution for a point-like source at x’.
By superposition, we can get a solution for source j.
 ( x)   ( x)   d x'G ( x, x' ) j( x' )
0
4
Green Function for KG Equation:

   G ( x, x' )  m  ( x, x' )  
2
4
 x  x '
By translation invariance, G is only a function of coordinate
difference:
G ( x, x' )  G ( x  x' )
The Equation becomes algebraic after a Fourier transformation.
4
G ( x  x') 
d p
 2 
4
d p
  2 
 p
4
2
e
 ip  ( x  x ' )
m
~
G ( p) 
2


4

~
G ( p)
~
p  m G ( p) 
2
2
4
 ( x  x') 
4
p m
2
d p
  2 
4
d p
  2 
4
e
 ip  ( x  x ' )
~
G ( p)  1
1
2
e
 ip  ( x  x ' )
This is the propagator!
4
e
 ip  ( x  x ' )
KG Propagation
x'
x
Green function is the effect at x of a source at x’.
That is exactly what is represented in this diagram.
The tricky part is actually the boundary condition.
For those amplitude where
time 2 is ahead of time 1,
propagation is from 2 to 1.
For those amplitude where
time 1 is ahead of time 2,
propagation is from 1 to 2.
B
B
B
2
B
1
1
C
A
2
C
A
A
A
is actually the sum of the above two diagrams!
~
To accomplish this, G ( p ) 
1
p m
2
2
~
G ( p) 
1
p  m  i
2
2